Read the passage carefully and answer the questions.
Amines are considered to be derivatives of ammonia obtained by replacement of hydrogen atoms with alkyl or aryl groups. They are usually prepared from nitro compounds, halides, amides, imides etc. The presence of hydrogen bonding influences its physical properties. In alkylamines, a combination of electron releasing steric and hydrogen bonding factors influences the stability of substituted ammonium cations in protic polar solvents and thus affects the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia which is more basic than water. Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding. This intermolecular association is more in primary amines than in secondary amines. Tertiary amines do not have intermoleulcar association due to the absence of hydrogen atom available for hydrogen bonding. Thus, the boiling point of amines depends on hydrogen bonding. In aromatic amines, electron releasing and withdrawing groups increase and decrease their basic character. Aryl diazonium salts usually obtained from arylamine undergo replacement of the diazonium group with a variety of nucleophile. Their coupling reaction of aryl diazonium salts with phenol or arylamine results in formation of dyes.
Which of the following compounds will not undergo azo coupling with benzene diazonium chloride?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Nitrobenzene
In azo coupling the benzene diazonium ion acts as an electrophile and attacks another aromatic ring, usually at the para position, joining the two rings through –N=N–. The diazonium ion is a weak electrophile, so it can attack only a ring made electron-rich by a strongly activating group such as –OH, –\( \ce{NH2} \) or –\( \ce{OCH3} \).
Option (1) Anisole — Undergoes coupling (activated ring). The –\( \ce{OCH3} \) group activates the ring by resonance and directs to the para position, so anisole couples to give p-methoxyazobenzene, though more slowly than phenol.
Option (2) Phenol — Undergoes coupling. Phenol couples readily in mildly alkaline solution, where the phenoxide ion is even more strongly activated, giving orange p-hydroxyazobenzene.
\( \displaystyle \ce{C6H5N2^+Cl^- + C6H5OH} \xrightarrow{\ce{OH-}} p\text{-}\ce{HO-C6H4-N=N-C6H5 + HCl} \)
Option (3) Aniline — Undergoes coupling. The –\( \ce{NH2} \) group strongly activates the ring, and coupling in mildly acidic solution gives yellow p-aminoazobenzene.
\( \displaystyle \ce{C6H5N2^+Cl^- + C6H5NH2} \xrightarrow{\ce{H+}} p\text{-}\ce{H2N-C6H4-N=N-C6H5 + HCl} \)
Option (4) Nitrobenzene — Does not undergo coupling (deactivated ring). The –\( \ce{NO2} \) group withdraws electrons from the ring by both inductive and resonance effects and deactivates it, so the weak diazonium electrophile cannot attack it.
NCERT (2026-27 edition):
Reactions Involving Retention of Diazo Group
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | 23-05 S2 13-05 S1 | – | 15-06 S1 31-05 S2 | – |